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\n" ); document.write( "\n" ); document.write( "some rationalisation, surds? Not sure how to do this. Any help appreciated.
Algebra.Com's Answer #650997 by Theo(13342)![\"\"](\"/images/stars/stars-13.gif\")  You can put this solution on YOUR website! your equation is 3-5x = square root of 5, presumably. \n" ); document.write( "square both sides of the equation to get (3-5x)^2 = 5 \n" ); document.write( "simplify to get 9 - 30x + 25x^2 = 5 \n" ); document.write( "subtract 5 from both sides of the equation to get 4 - 30x + 25x^2 = 0 \n" ); document.write( "reorder the terms in descending order of degree to get 25x^2 - 20x + 4 = 0 \n" ); document.write( "factor this quadratic equation to get x = .1527864045 or x = 1.047213596. \n" ); document.write( "when you replace x, in the original equation of 3 - 5x = sqrt(5), with .1527864045, you get 3 - 5*.1527864045 = 2.236067977, which results in 2.236067977 = 2.236067977 which is true. \n" ); document.write( "when you replace x, in the original expression of 3 - 5x, with 1.047213596, you get 3 - 5 * 1.047213596 = -2.236067978, which results in -2.236067977 = 2.236067977 which is not true. \n" ); document.write( "since sqrt(5) = 2.236067977 only, this means that your solution is x = .1527864045 only. \n" ); document.write( "you need to discard x = 1.047213596 because, even though it is a solution to the quadratic equation of (3 - 5x)^2 = 5, it is not a solution to the original equation of 3 - 5x = sqrt(5. \n" ); document.write( " \n" ); document.write( " |